Solution 8 to problem N1t8s12f1

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t8s12f1

Expressions

The solution is given through the following expressions: 

q35=0


q34=0


q33=0


q32=0


q31=0


q30=0


q29=0


q28=0


q27= - q3


q26= - 6*q3


q25=0


q24=0


q23=0


q22=0


q21=0


q20=0


q19=0


q18=0


q17=0


q16=2*q3


q15=0


q14=0


q13= - q3


        1
q12= - ---*q9
        5


q11=0


q10=0


q8=q3


q7= - 2*q3


q6=0


q5=0


q4=0


q2=0


q1=0


p13=0


p12=0


p11=0


p10=0


p9=0


p8=0


p7=0


       1
p6= - ---*p4
       3


p5=0


p3=0


p2=0


p1=0

Parameters

Apart from the condition that they must not vanish to give a non-trivial solution and a non-singular solution with non-vanishing denominators, the following parameters are free: 
 q9, q3, p4

Inequalities

In the following not identically vanishing expressions are shown. Any auxiliary variables g00?? are used to express that at least one of their coefficients must not vanish, e.g. g0019*p4 + g0020*p3 means that either p4 or p3 or both are non-vanishing. 
 
{q3,

 p4,

 15*g0012*q3 + 90*g0013*q3 - 30*g0023*q3 + 15*g0026*q3 + 3*g0027*q9

  - 15*g0030*q9 - 15*g0031*q3 + 30*g0032*q3 - 15*g0036*q3 + 5*g0045*p4

  - 15*g0047*p4,

 p6,

 5*g0059*q3 + 30*g0060*q3 - 10*g0070*q3 + 5*g0073*q3 + g0074*q9 - 5*g0077*q9

  - 5*g0078*q3 + 10*g0079*q3 - 5*g0083*q3}

Relevance for the application:



The equation: 



           2         1      3
f =Df *(Df) *f*p4 - ---*(Df) *f *p4
 t   x               3         x


The symmetry: 

            2                                        2              3
f =Df  *(Df) *f*q3 - 2*Df  *Df *Df*f*q3 + 2*Df  *(Df) *f *q3 + (Df ) *f*q3
 s   3x                  2x   x               2x        x         x

        2                    4         1      5             3
 - (Df ) *Df*f *q3 + Df *(Df) *f*q9 - ---*(Df) *f *q9 - (Df) *f  *q3
      x       x        x               5         x             3x

 - 6*Df*f  *f *f*q3
         3x  x


And now in machine readable form: 

The system: 

df(f(1),t)=d(1,df(f(1),x))*d(1,f(1))**2*f(1)*p4 - 1/3*d(1,f(1))**3*df(f(1),x)*p4
$


The symmetry: 

df(f(1),s)=d(1,df(f(1),x,3))*d(1,f(1))**2*f(1)*q3 - 2*d(1,df(f(1),x,2))*d(1,df(f
(1),x))*d(1,f(1))*f(1)*q3 + 2*d(1,df(f(1),x,2))*d(1,f(1))**2*df(f(1),x)*q3 + d(1
,df(f(1),x))**3*f(1)*q3 - d(1,df(f(1),x))**2*d(1,f(1))*df(f(1),x)*q3 + d(1,df(f(
1),x))*d(1,f(1))**4*f(1)*q9 - 1/5*d(1,f(1))**5*df(f(1),x)*q9 - d(1,f(1))**3*df(f
(1),x,3)*q3 - 6*d(1,f(1))*df(f(1),x,3)*df(f(1),x)*f(1)*q3$